Question

A $30kg$ box has to move up an inclined slope of ${30}^{\circ }$ to the horizontal at a uniform velocity of $5m{s}^{-1}$. If the frictional force retarding the motion is $150N$, the horizontal force required to move up is $(g=10m{s}^{-2})$.

Answer 1

$F-f=mg\sin \theta$
$F=mg\sin \theta +f$
$=30\times 10\times \frac{1}{2}+150$
$=300N$
$F^{\prime }\cos 30=F$
$F^{\prime }=\frac{F}{\cos 30}=\frac{300}{\frac{\sqrt{3}}{2}}$.